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我希望看到发送请求数据包之前,我发送它作为req中有一个错误和api是一般性描述500 错误,所以我不能告诉y请求失败。我知道xml的格式不正确,因为它在Chrome上运行在postman上。查看guzzle发帖请求
$client = new GuzzleHttp\Client([
'base_uri' => 'https://elstestserver.endicia.com',
]);
$xml = 'changePassPhraseRequestXML=<ChangePassPhraseRequest> <RequesterID>lxxx</RequesterID><RequestID>1263055835</RequestID><CertifiedIntermediary><AccountID>lxxx</AccountID><PassPhrase>dfdsfsd</PassPhrase></CertifiedIntermediary><NewPassPhrase>fdfdsfdsfs</NewPassPhrase></ChangePassPhraseRequest>';
$data = array("ChangePassPhraseXML" => $xml);
$response = $client->post("/LabelService/EwsLabelService.asmx/ChangePassPhraseXML", [
'form_params' => $data
]);
申请工作在邮递员铬的继承人的XML的工作示例
changePassPhraseRequestXML=<ChangePassPhraseRequest><RequesterID>lxxx</RequesterID><RequestID>1263055835</RequestID><CertifiedIntermediary><AccountID>lxxx</AccountID><PassPhrase>dfdsfsd</PassPhrase></CertifiedIntermediary><NewPassPhrase>fdsfdsfds</NewPassPhrase></ChangePassPhraseRequest>
你会寻找调试请求选项。如果在客户端实例化中将其设置为true,则客户端发出的所有请求都会将请求输出到php:// stdout。供参考:http://guzzle.readthedocs.org/en/latest/request-options.html#debug –